HowTo: reg expr doing grep "timestamp>$DesiredTime" logfile ?

Browser_ice · #1 (**permalink**) 06-25-2009

I know I asked a similar question but I want to know if there is a regular expression existing that with a korn shell cmd, finds any timestamp data records in a file where it is greater then a timestamp in a shell variable ?

something like :

grep all records where it has a timestamp > $DesiredTime from logfile-X

It would simplify my coding alot for what I want to do on the client's AIX 4.2

Browser_ice · #2 (**permalink**) 07-04-2009

Any ideas ?

Franklin52 · #3 (**permalink**) 07-04-2009

How does the format of the dates lookes like in your log file?

Browser_ice · #4 (**permalink**) 07-05-2009

Quote:

Originally Posted by Franklin52

How does the format of the dates lookes like in your log file?

like the standard `date` output:

I know I could always do a regular expression to match each date parts. But I need to know if there a special one that will allow me something like :
grep all records in file where any timestamp in it is > $a_given_timestamp

vidyadhar85 · #5 (**permalink**) 07-05-2009

Quote:

Originally Posted by Browser_ice

like the standard `date` output:

I know I could always do a regular expression to match each date parts. But I need to know if there a special one that will allow me something like :
grep all records in file where any timestamp in it is > $a_given_timestamp

For that I afraid you have to convert your timestamp to min or sec's before comapring it with the given timestamp(that should be also in min or in sec's)

Franklin52 · #6 (**permalink**) 07-05-2009

To match dates you can use a format like YYYYMMDDHH, an example to convert the date format Mar 17 16:44:03 2009 to 2009031716

Code:

echo "Mar 17 16:44:03 2009"|
awk -F" |:" '
BEGIN {
  a="Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec"
  split(a,m,",")
  for(i=1;i<13;i++)
    n[m[i]]=i
}
{printf("%s%02d%02d%s\n", $6, n[$1], $2, $3)}
'

Use nawk or /usr/xpg4/bin/awk on Solaris if you get errors.

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